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A voltage commutation circuit is shown in figure. The turn-off time of the both SCR is 60 sec and current through R1 and R2 is 20 A and voltage is 120 V. What will be the minimum value of capacitor (in μF) required for proper commutation? (Answer up to one decimal place)


Correct answer is '14.4'. Can you explain this answer?
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Given information:
- Turn-off time of both SCR is 60 sec.
- Current through R1 and R2 is 20 A.
- Voltage across R1 and R2 is 120 V.

To determine:
The minimum value of the capacitor required for proper commutation.

Approach:
- The commutation process in a voltage commutation circuit involves turning off one SCR while turning on the other SCR to ensure continuous flow of current.
- The minimum value of the capacitor required for proper commutation can be determined by considering the worst-case scenario, where the turn-off time of both SCRs is at its maximum value.

Calculation:
- The worst-case scenario occurs when both SCRs turn off simultaneously.
- The voltage across the capacitor during this worst-case scenario is given by the equation: Vc = V - Vr2 - Vr1, where V is the supply voltage, Vr2 is the voltage across resistor R2, and Vr1 is the voltage across resistor R1.
- V = 120 V (given)
- Vr1 = I * R1 = 20 A * 10 Ω = 200 V
- Vr2 = I * R2 = 20 A * 5 Ω = 100 V
- Vc = V - Vr2 - Vr1 = 120 V - 100 V - 200 V = -180 V (negative because the polarities are reversed during commutation)
- The voltage across the capacitor cannot be negative, so we need to choose a minimum value of the capacitor such that Vc is positive.
- Let C be the capacitance in Farads.
- The charge stored in the capacitor during commutation is given by the equation: Q = C * Vc.
- For proper commutation, the charge stored in the capacitor should be sufficient to maintain the voltage across the load during the turn-off time of the SCR.
- Let Toff be the turn-off time of the SCR = 60 sec.
- The charge required to maintain the voltage across the load is given by the equation: Qload = I * Toff.
- Equating the charge stored in the capacitor to the charge required to maintain the voltage across the load, we have: C * Vc = Qload.
- Substituting the values, we get: C * (-180 V) = 20 A * 60 sec.
- Solving for C, we get: C = 14.4 μF.

Therefore, the minimum value of the capacitor required for proper commutation is 14.4 μF.
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A voltage commutation circuit is shown in figure. The turn-off time of the both SCR is 60 sec and current through R1 and R2 is 20 A and voltage is 120 V. What will be the minimum value of capacitor (in μF) required for proper commutation? (Answer up to one decimal place)Correct answer is '14.4'. Can you explain this answer?
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A voltage commutation circuit is shown in figure. The turn-off time of the both SCR is 60 sec and current through R1 and R2 is 20 A and voltage is 120 V. What will be the minimum value of capacitor (in μF) required for proper commutation? (Answer up to one decimal place)Correct answer is '14.4'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A voltage commutation circuit is shown in figure. The turn-off time of the both SCR is 60 sec and current through R1 and R2 is 20 A and voltage is 120 V. What will be the minimum value of capacitor (in μF) required for proper commutation? (Answer up to one decimal place)Correct answer is '14.4'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A voltage commutation circuit is shown in figure. The turn-off time of the both SCR is 60 sec and current through R1 and R2 is 20 A and voltage is 120 V. What will be the minimum value of capacitor (in μF) required for proper commutation? (Answer up to one decimal place)Correct answer is '14.4'. Can you explain this answer?.
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